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Let ABC be a triangle with incentre I an...

Let ABC be a triangle with incentre I and inradius r. Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA and AB, respectively, If `r_(2)" and "r_(3)` are the radii of circles inscribed in the quadrilaterls AFIE, BDIF and CEID respectively, then prove that `r_(1)/(r-r_(1))+r_(2)/(r-r_(2))+r_(3)/(r-r_(3))=(r_(1)r_(2)r_(3))/((r-r_(1))(r-r_(2))(r-r_(3)))`

Text Solution

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I is incenter of `Delta ABC`.
`r_(1)` is the radius of the circle inscribed in quadrilateral AFIE. From the figure, AF and AE are tangents to the circle.
Also IE and IF are tangent to the circle.
So AI is passing through the center of the circle by summetry.
Now ON is parallel to IE
So `Delta ANO and Delta OMI` are similar.
MI, we have `cot.(A)/(2) = (OM)/(IM) = (r_(1))/(r - r_(1))`
Similarly from other circles we get
`cot. (B)/(2) = (r_(2))/(r - r_(2)) and cot. (C)/(2) = (r_(3))/(r - r_(3))`
Now in `Delta ABC`, we know that
`cot.(A)/(2) cot.(B)/(2) cot.(C)/(2) = cot.(A)/(2) + cot.(B)/(2) + cot.(C)/(2)`
`(r_(1))/(r - r_(1)).(r_(2))/(r - r_(2)) . (r_(3))/(r - r_(3)) = (r_(1))/(r - r_(1)) + (r_(2))/(r - r_(2)) + (r_(3))/(r - r_(3))`
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